Home Physics Newton's Laws of Motion First Law of Motion A pearl of mass m is in a position to slide …
Physics Newton's Laws of Motion First Law of Motion MCQ (Single Correct)

A pearl of mass m is in a position to slide over a smooth wire. At the initial instant the pearl is in the middle of the wire. The wire moves linearly in a horizontal plane with an acceleration a in a direction having angle θ θ with the wire. The acceleration of the pearl with reference to wire is

A
g sin θ θ - a cos θ θ
B
g sin θ θ - g cos θ θ
C
g sin θ θ + a cos θ θ
D
g cos θ θ + a sin θ θ

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Text Solution

Verified by Experts
The correct answer is:
A

Let a x , a y and a r be the net leftward horizontal acceleration bead, net downward vertical acceleration of bead and relative acceleration of bead with reference to rod respectively. Then

a y = a r cos θ θ + a

and a x = a r sin θ θ

Projecting forces vertically and horizontally

mg – N cos θ θ = ma r + ma cos θ θ

i.e., a r = g sin θ θ - a cos θ θ

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